As it stands, intsin^2xdx has no clear solutions. This is one of those cases where knowing your algebra and trig comes in real handy.
You may recall that sin^2x=1/2(1-cos2x) (if you don't, that's fine because I forgot too). So, instead of integrating sin^2x, we can integrate 1/2(1-cos2x) and see if that gets us anywhere:
int1/2(1-cos2x)dx
=1/2int1-cos2xdx
We can use the sum rule to split this into:
1/2int1dx-1/2intcos2xdx
We're getting somewhere because the first of these integrals, int1dx, is the "perfect integral" and evaluates to x:
1/2x-1/2intcos2xdx
We just have to worry about intcos2xdx. Let's start off with what we know:intcosxdx=sinx because the derivative of sinx is cosx. We just have to adjust for that pesky 2.
Let's think for a moment. intcos2xdx essentially means that if we take the derivative of our solution, we should get cos2x. Let's guess a solution of 1/2sin2x and see what happens when we differentiate it:
d/dx1/2sin2x=1/2*(2x)'*cos2x=1/2*2*cos2x=cos2x
Since that solution checks out, we can say that intcos2xdx=1/2sin2x
Ok, so we have 1/2x-1/2intcos2xdx. We just found the last integral, so our solution is 1/2x-1/2(1/2sin2x)+C=1/2x-1/4sin2x+C (don't forget the integration constant C!)
